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CBSE NCERT Solutions for Class 11 Chemistry Chapter 12
Back of Chapter Questions
1. What are hybridisation states of each carbon atom in the following compounds?
CH2 = C = O, CH3CH = CH2, (CH3)2CO, CH2 = CHCN, C6H6
Solution:
Hybridisation = σ bond + lone pair
If 2σ=sp
3σ = sp2
4σ = sp3
(i)
C − 1 is sp2 hybridised
C − 2 is sp hybridized
(ii)
C − 3 is sp3 hybridised
C − 2 is sp2 hybridised
C − 1 is sp2 hybridised
(iii)
C − 1 and C − 3 are sp3 hybridised
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C − 2 is sp2 hybridised
(iv)
C − 3 is sp2 hybridised
C − 2 is sp2 hybridised
C − 1 is sp hybridized
(v) C6H6
All the 6 carbon atoms in benzene are sp2 hybridised.
2. Indicate the σ and π bonds in the following molecules:
C6H6, C6H12, CH2Cl2, CH2 = C = CH2, CH3NO2, HCONHCH3
Solution:
(i) C6H6
There are six C – C sigma (σC−C) bonds, six C–H sigma (σC−H) bonds, and three C = C
pi
(πC−C) resonating bonds in the given compound.
(ii) C6H12
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There are six C − C sigma (σC−C) bonds and twelve C − H sigma (σC−H) bonds in the
given compound.
(iii) CH2Cl2
There two C − H sigma (σC−H) bonds and two C − Cl sigma (σC−Cl) bonds in the given
compound.
(iv) CH2 = C = CH2
There are two C–C sigma (σC−C) bonds, four C–H sigma (σC−H) bonds, and two C = C
pi
(πC−C) bonds in the given compound.
(v) CH3NO2
There are three C–H sigma (σC−H) bonds, one C–N sigma (σC−N) bond, one N–O sigma
(σN−O) bond, and one N = O pi (πN−O) bond in the given compound.
(vi) HCONHCH3
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There are two C–N sigma (σC−N) bonds, four C–H sigma (σC−H) bonds, one N–H sigma
(σN−H) bond, and one C = O pi (πC−O) bond in the given compound.
3. Write bond-line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
Solution :
(a) Isopropyl alcohol
formulas of Isopropyl alcohol :
bond-line formulas of Isopropyl alcohol:
(b) 2, 3–dimethyl butanal
formula of 2,3-Dimethylbutanal:
bond line formula of 2,3-Dimethylbutanal:
(C) Heptan–4–one
formulas for: Heptan-4-one:
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bond line formulas of Heptan-4-one.
4. Give the IUPAC names of the following compounds:
(A)
(B)
(C)
(D)
(E)
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(F) Cl2CHCH2OH
Solution:
(A)
3-phenyl propane
(B)
2-methyl-1-cyanobutane
(C)
2, 5- dimethyl heptane
(D)
3-bromo-3-chloroheptane
(E)
3- chloropropanal
(F)
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2, 2-dichloroethan-1-ol
5. Which of the following represents the correct IUPAC name for the compounds
concerned?
(A) 2,2-Dimethylpentane or 2-Dimethylpentane
(B) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane
(C) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
(D) But-3-yn-1-ol or But-4-ol-1-yne
Solution:
(A) The prefix di in the IUPAC name indicates that two identical substituent (−CH3) are
present in the parent chain. Since two methyl groups are present in the C − 2 of the parent
carbon chain of the given compound, the correct IPUAC name of the given compound is
2, 2-dimethylpentane.
(b)Locant number 2, 4, 7 is lower than 2, 5, 7. Hence, the IUPAC name of the given
compound is 2, 4, 7-trimethyloctane.
(c) If the substituents are present in the equivalent position of the parent chain, then the
lower number is given to the one that comes first in the name according to the
alphabetical order and priorler group. Hence, the correct IUPAC name of the given
compound is 2-chloro- 4-methylpentane.
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(d) Two functional groups − alcoholic and alkyne − are present in the given compound.
The principal functional group is the alcoholic group. Hence, the parent chain will be
suffixed with ol and get least number. The alkyne group is present in the C − 3 of the
parent chain. Hence, the correct IUPAC name of the given compound is But−3 − yn −l − ol.
6. Draw formulas for the first five members of each homologous series beginning with
the following compounds.
(A) H–COOH
(B) CH3COCH3
(C) H–CH = CH2
Solution:
Homologous series : the series of compounds which have a difference of - CH2 or mass
of 14 u.and the functional group should remain the same
The first five members of each homologous series beginning with the given compounds
are shown as follows:
(a)
H–COOH : Methanoic acid
CH3 − COOH : Ethanoic acid
CH3 − CH2 − COOH : Propanoic acid
CH3 − CH2 − CH2 − COOH : Butanoic acid
CH3 − CH2 − CH2 − CH2 − COOH : Pentanoic acid
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(b)
CH3COCH3 : Propanone
CH3COCH2CH3: Butanone
CH3COCH2CH2CH3 : Pentan-2-one
CH3COCH2CH2CH2CH3 : Hexan-2-one
CH3COCH2CH2CH2CH2CH3 : Heptan-2-one
(c)
H − CH = CH2 : Ethene
CH3 − CH = CH2 : Prop-1-ene
CH3 − CH2 − CH = CH2 : But-1-ene
CH3 − CH2 − CH2 − CH = CH2 : Pent-1-ene
CH3 − CH2 − CH2 − CH2 − CH = CH2 : Hex-1-ene
7. Give condensed and bond line structural formulas and identify the functional group(s)
present, if any, for:
(A) 2,2,4-Trimethylpentane
(B) 2-Hydroxy-1,2,3-propanetricarboxylic acid
(C) Hexanedial
Solution:
(a) 2, 2, 4–trimethylpentane
Condensed formula : (CH3)2CHCH2C(CH3)3
Bond line formula:
(b) 2–hydroxy–1, 2, 3–propanetricarboxylic acid
Condensed Formula: (COOH)CH2C(OH)(COOH)CH2(COOH)
Bond line formula:
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The functional groups present in the given compound are carboxylic acid (−COOH) and
alcoholic (−OH) groups.
(c) Hexanedial
Condensed Formula :
(CHO)(CH2)4(CHO)
Bond line Formula :
The functional group present in the given compound is aldehyde (−CHO).
8. Identify the functional groups in the following compounds
(A)
(B)
(C)
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Solution:
The functional groups present in the given compounds are:
(a)
Aldehyde (−CHO), Hydroxyl (−OH),
Methoxy (−OMe),
C = C double bond (−C1
= C1
−)
(b)
Amino (−NH2),
Ester: RCOOR
Tertiary amine :(R)3N
(c)
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Nitro (−NO2),
C = C double bond (alkene) (−C = C −)
9. Which of the two: O2NCH2CH2O– or CH3CH2O
– is expected to be more stable and why?
Solution:
-NO2 group is an electron-withdrawing group. It shows – I effect. By withdrawing the
electrons toward it, the - NO2 group decreases the negative charge density on the
compound, thereby stabilising the negative charge . On the other hand, ethyl group is an
electron-releasing group. It shows +I effect. This increases the negative charge on the
compound, thereby destabilising the anion. Hence, O2NCH2CH2O− is more stable than
CH3CH2O–.
10. Explain why alkyl groups act as electron donors when attached to a π system.
Solution:
When an alkyl group is attached to a π system, it acts as an electron-donor group by the
process of hyperconjugation. To understand this concept better, let us take the example
of propene.
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In hyperconjugation, the sigma electrons of the C − H bond of an alkyl group are
delocalised. This group Is directly attached to an atom of an unsaturated system. The
delocalisation occurs because of a partial overlap of a sp3 − s sigma bond orbital with an
empty p orbital of the π bond of an adjacent carbon atom.
The process of hyperconjugation in propene is shown as follows:
This type of overlap leads to a delocalisation (also known as no-bond resonance) of the π
electrons, making the molecule more stable.
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11. Draw the resonance structures for the following compounds. Show the electron shift
using curved-arrow notation.
(A) C6H5OH
(B) C6H5NO2
(C) CH3CH = CHCHO
(D) C6H5– CHO
(e) C6H5 − C+
H2
(f) CH3CH = CHC+
H2
Solution:
(a) The structure of C6H5OH is:
The resonating structures of phenol are represented as:
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(B) The structure of C6H5NO2 is:
The resonating structures of nitro benzene are represented as:
(c) CH3CH = CH − CHO
The resonating structures of the given compound are represented as:
(D) The structure of C6H5CHO is:
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The resonating structures of benzaldehyde are represented as:
(E) C6H5CH2⊕
The resonating structures of the given compound are:
(F) CH3CH = CHCH2⊕
The resonating structures of the given compound are:
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12. What are electrophiles and nucleophiles? Explain with examples.
Solution:
Electro means electron and phile means loving so electrophile si electron loving. In other
words, an electron-deficient reagent is called an electrophile (E+).since Electrophiles are
electron deficient so they can accept electron pair. Some positively charged species
(carbocation) and octet deficient species are electrophiles
examples of electrophiles.
Carbocations (CH3CH2+) , Cl+, Br+, I+ has 6 electrons in valance shell
neutral molecules like BF3, AlCl3 has 6 electrons in valance shell
functional groups such as carbonyl group behave like electrophile.
nucleophile (Nu: ) : Nucleo means nucleus and phile means loving so nucleophile is
nucleus loving. In other words, an electron-rich reagent is called a nucleophile (nu−).A
nucleophile is a reagent that brings an electron pair or negative charge density.
For example negatively charged species like OH−, NC−, carbanions
R3C−),F−, Cl−, I−, Br−, N3
−, etc.
Neutral molecules such as H2O··
and ammonia also acts as nucleophiles because of the
presence of a lone pair.
13. Identify the reagents shown in bold in the following equations as nucleophiles or
electrophiles:
(A) CH3COOH + HO− → CH3COO
− + H2O
(B) CH3COCH3 + C−
N → (CH3)2C(CN)(OH)
(C) C6H6 + CH3C+
= O → C6H5COCH3
Solution:
Electrophiles are electron-deficient species and can receive an electron pair. On the other
hand, nucleophiles are electron-rich species and can donate their electrons.
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(A) CH3COOH + HO− → CH3COO
− + H2O
Here, HO− acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus
seeking
species.
(B) CH3COCH3 + C̄N → (CH3)2C(CN) + (OH)
Here, −CN acts as a nucleophile as it is an electron-rich species, it will attack at nucleus
or electron-deficient side
(C) C6H5 + CH3C+
O → C6H5COCH3
Here, CH3C+
O acts as an electrophile as it is an electron-deficient species or it attack at
electron rich side .
14. Classify the following reactions in one of the reaction type studied in this unit.
(A) CH3CH2Br + HS− → CH3CH2SH + Br
−
(B) (CH3)2C = CH2 + HCl → (CH3)2ClC − CH3
(C) CH3CH2Br + HO− → CH2 = CH2 + H2O + Br
−
(D) (CH3)3C − CH2OH+ HBr → (CH3)2CBrCH2CH2CH3 + H2O
Solution:
(A) CH3CH2Br + HS− → CH3CH2SH + Br
− It is an example of nucleophilic substitution reaction as in this reaction the
bromine group in is leaving group and -Br group is substituted by the −SH
group.
(B) (CH3)2C = CH2 +HCl → (CH3)2ClC − CH3 It is an example of electrophilic addition reaction as in this reaction two reactant
molecules combine to form product.
(C) CH3CH2Br + HO− → CH2 = CH2 + H2O + Br
− It is an example of elimination reaction as in this reaction hydrogen and bromine
are removed from bromoethane to give ethene.
(D) (CH3)3C − CH2OH+ HBr → (CH3)2CBrCH2CH2CH3 + H2O
In this reaction, substitution takes place, -OH group is replaced with -Br group and it
followed by a rearrangement of atoms and groups of atoms.
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15. What is the relationship between the members of following pairs of structures? Are they
structural or geometrical isomers or resonance contributors?
(A)
(B)
(C)
Solution:
(A) Compounds having the same molecular formula but with different structures are
called structural isomers. The given compounds have the same function group (ketone)
but they differ in the position.
(B) Compounds having the same molecular formula, the same connectivity of atoms and
their arrangement in space is different .in the given compounds the restricted rotation
present du to double bond and different groups attached with carbon, so they are called
geometrical isomers.
Priority of group D > H
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In structures I and II, the relative position of Deuterium (D) and hydrogen (H) in space
are different. Hence, the given pairs represent geometrical isomers.
(C) The given structures are canonical structures or contributing structures. They are
hypothetical and individually do not represent any real molecule. Hence, the given pair
represents resonance structures, called resonance isomers.
16. For the following bond cleavages, use curved-arrows to show the electron flow and
classify each as homolysis or heterolysis. Identify reactive intermediate produced as free
radical, carbocation and carbanion.
(A)
(B)
(C)
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(D)
Solution:
(a)
It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes
with the bonded atom because there is no electronegativity difference . The reaction
intermediate formed is a free radical.
The bond cleavage using half curved-arrows to show the electron flow
free radical
(B)
In this reaction base (OH−) snatched the acidic hydrogen . acidic hydrogen leave as
(H+ )
There is an electronegativity difference between carbon and hydrogen so it is an example
of heterolytic cleavage as the bond breaks in such a manner that the shared pair of
electrons remains with the carbon of propanone. The reaction intermediate formed is
carbanion
The bond cleavage using curved-arrows to show the electron flow of the given reaction
can be represented as
(C)
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It is an example of heterolytic cleavage because of electronegativity difference in
between Carbon and Bromine.as the bond breaks in such a manner that the shared pair of
electrons remains with the bromine ion and carbon has electron deficiency so reaction
intermediate formed is a carbocation.
The bond cleavage using curved-arrows to show the electron flow of the given reaction
(D)
It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of
electrons remains with one of the fragments. The intermediate formed is a carbocation.
17. Explain the terms Inductive and Electromeric effects. Which electron displacement effect
explains the following correct orders of acidity of the carboxylic acids?
(A) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH
(B) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C. COOH
Solution:
Inductive effect is a permanent displacement of sigma (σ) electrons along a saturated
chain, whenever an electron-withdrawing or electron-donating group is present, is called
the inductive effect. It occurs due to electronegativity of the difference of atoms
It is a feeble effect and as the distance increases the strength of the inductive effect
decreases
Inductive effect ∝1
distance
The inductive effect is two types +I effect or −I effect.
−I effect : When an atom or group attract electrons towards itself more strongly than
hydrogen
For example,
+I effect : When an atom or group attract electrons towards itself less strongly than
hydrogen .
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For example,
Electrometric effect
It involves the complete transfer of the shared pair of π electrons to either of the two
atoms linked by multiple bonds in the presence of an attacking agent. For example,
Electrometric effect could be +E effect or −E effect.
+E effect: When the electrons are transferred towards the attacking reagent
−E effect: When the electrons are transferred away from the attacking reagent
(a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH
Acidic strength ∝ stability of conjugate base (Anion)
Stability of conjugate base or anion ∝1
electron density
The order of acidity can be explained on the basis of Inductive effect (−I effect). As the
number of chlorine atoms increases, the −I effect increases. With the increase in −I effect, the acid strength also increases accordingly.
Conjugate base stability or acidic strength order
Electron density order.
(B) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C. COOH
Acidic strength ∝ stability of conjugate base (Anion)
Stability of conjugate base or anion ∝1
electron density
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The order of acidity can be explained on the basis of inductive effect (+I effect). As the
number of alkyl groups increases, the +I effect also increases. With the increase in +I effect, the acid strength also increases accordingly.
Stability of anion or conjugate base order.
Electron density order.
18. Give a brief description of the principles of the following techniques taking an example
in each case.
(A) Crystallisation
(B) Distillation
(C) Chromatography
Solution:
(a) Crystallisation
Crystallisation techniques is used for the purification of solid organic compounds.
Principle: used : It is based on the difference in the solubilities of the compound and the
impurities in a suitable solvent. The impure compound is dissolved in a solvent in which
it is sparingly soluble at room temperature but appreciably soluble at higher temperature.
The solution is concentrated to get a nearly saturated solution. On cooling the solution,
pure compound crystallises out and is removed by filtration.
The filtrate (mother liquor) contains impurities and small quantity of the compound. If
the compound is highly soluble in one solvent and very little soluble in another solvent,
crystallisation can be satisfactorily carried out in a mixture of these solvents. Impurities,
which impart colour to the solution are removed by adsorbing over activated charcoal.
Repeated crystallization becomes necessary for the purification of compounds containing
impurities of comparable solubilities.
For example, pure aspirin is obtained by recrystallising crude aspirin. Approximately 2 –
4 g of crude aspirin is dissolved in about 20 mL of ethyl alcohol. The solution is heated
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(if necessary) to ensure complete dissolution. The solution is then left undisturbed until
some crystals start to separate out. The crystals are then filtered and dried.
(B) Distillation : used to separate volatile liquids from non-volatile impurities or a
mixture
of those liquids that have a sufficient difference in their boiling points.
Principle: It is based on the fact that liquids having different boiling points vapourise at
different temperatures. The vapours are then cooled and the liquids so formed are
collected separately.
For example, a mixture of chloroform (b. p = 334 K) and aniline (b. p = 457 K) can be
separated by the method of distillation. The mixture is taken in a round bottom flask
fitted
with a condenser. It is then heated. Chloroform, being more volatile, vaporizes first and
passes into the condenser. In the condenser, the vapours condense and chloroform trickles
down. In the round bottom flask, aniline is left behind.
(C) Chromatography : useful methods for the separation and purification of organic
compounds.
Principle: It is based on the difference in movement of individual components of a
mixture
through the stationary phase under the influence of mobile phase.
For example, a mixture of red and blue ink can be separated by chromatography. A drop
of the mixture is placed on the chromatogram. The component of the ink, which is less
adsorbed on the chromatogram, moves with the mobile phase while the less adsorbed
component remains almost stationary.
19. Describe the method, which can be used to separate two compounds with different
solubilities in a solvent.
Solution:
Fractional crystallisation is the method used for separating two compounds with different
solubilities in a solvent.
The process of fractional crystallisation is carried out in four steps.
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Step 1- Preparation of the solution: The powdered mixture is taken in a flask and the
solvent is added to it slowly and stirred simultaneously. The solvent is added till the
solute is just dissolved in the solvent. This saturated solution is then heated.
Step 2- Filtration of the solution: The hot saturated solution is then filtered through a
filter paper in a China dish.
Step 3- Fractional crystallisation: The solution in the China dish is now allowed to cool.
The less soluble compound crystallises first, while the more soluble compound remains
in the solution. After separating these crystals from the mother liquor, the latter is
concentrated once again. The hot solution is allowed to cool and consequently, the
crystals of the more soluble compound are obtained.
Step 4: Isolation and drying: These crystals are separated from the mother liquor by
filtration. Finally, the crystals are dried.
20. What is the difference between distillation, distillation under reduced pressure and steam
distillation?
Solution:
The differences among distillation, distillation under reduced pressure, and steam
distillation are given in the following table.
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Distillation Distillation under
reduced pressure
Steam distillation
1. It is used for the
purification of
compounds that are
associated with non-
volatile impurities or
those liquids, which do
not decompose on
boiling. In other words,
distillation is used to
separate volatile liquids
from non-volatile
impurities or a mixture
of those liquids that
have sufficient
difference in boiling
points.
This method is used to
purify a liquid that tends
to decompose on
boiling. Under the
conditions of reduced
pressure, the liquid will
boil at a low
temperature than its
boiling point and will,
therefore, not
decompose.
It is used to purify sn organic
compound, which is steam
volatile and immescible in
water. On passing steam, the
compound gets heated up and
the steam gets condensed to
water. After some time, the
mixture of water and liquid
starts to boil and passes
through the condenser. This
condensed mixture of water
and liquid is them separated
by using a separating funnel.
2. Mixture of petrol and
kerosene is separated by
this method.
Glycerol is purified by
this method. It boils
with decomposition at a
temperature of 593 K.
At a reduced pressure, it
boils at 453 K without
decomposition
A mixture of water and
aniline is separated by steam
distillation.
The differences among distillation, distillation under reduced pressure, and steam
distillation are given in the following table.
21. Discuss the chemistry of Lassaigne’s test.
Solution:
Lassaigne’s test
This test is used to detect the presence of nitrogen, sulphur, halogens, and phosphorous in
an organic compound. These elements are present in the covalent form in an organic
compound. These are converted into the ionic form by fusing the compound with sodium
metal.
Na + C + N ∆ → NaCN
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2Na + S ∆ → Na2S
Na + X ∆ → NaX
(X = Cl, Br, I)
The cyanide, sulphide, and halide of sodium formed are extracted from the fused mass by
boiling it in distilled water. The extract so obtained is called Lassaigne’s extract.
(a) Test for nitrogen
In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is
boiled with iron (II) sulphate and then acidified with concentrated sulphuric acid. In the
process, sodium
cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). which
is Prussian blue in colour Then,on heating with sulphuric acid, some iron (II) gets
oxidised to form iron (III)
(B) Test for sulphur
(i) Lassaigne’s extract + Lead acetate acetic acid→ Black precipitate.
In the Lassaigne’s test for sulphur in an organic compound, the sodium fusion extract is
acidified with acetic acid and then lead acetate is added to it. The precipitation of lead
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sulphide, which is black in colour, indicates the presence of sulphur in the compound.
or
Lassaigne’s extract + Sodium nitroprusside → Violet colour.
The sodium fusion extract is treated with sodium nitroprusside. Appearance of violet
colour
also indicates the presence of sulphur in the compound.
If in an organic compound, both nitrogen and sulphur are present, then instead of NaCN,
formation of NaSCN takes place.
Na + C + N + S → NaSCN
This NaSCN (sodium thiocyanate) gives a blood red colour. Prussian colour is not formed
due to the absence of free cyanide ions.
(C) Test for halogens
In the Lassaigne’s test for halogens in an organic compound, the sodium fusion extract is
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acidified with nitric acid and then treated with silver nitrate.
If nitrogen and sulphur both are present in the organic compound, then the Lassaigne’s
extract is boiled to expel nitrogen and sulphur, which would otherwise interfere in the test
for halogens.
22. Differentiate between the principle of estimation of nitrogen in an organic compound by
(i) Dumas method and
(ii) Kjeldahl’s method.
Solution:
In Dumas method, a known quantity of nitrogen containing organic compound is heated
strongly with excess of copper oxide in an atmosphere of carbon dioxide to produce free
nitrogen in addition to carbon dioxide and water. The chemical equation involved in the
process can be represented as
CxHyNz + (2x + y 2⁄ )CuO → xCO2 + y 2⁄ H2O+ z 2⁄ N2 + (2x + y 2⁄ )Cu
The traces of nitrogen oxides can also be produced in the reaction, which can be reduced
to dinitrogen by passing the gaseous mixture over a heated copper gauge. The dinitrogen
produced is collected over an aqueous solution of potassium hydroxide. The volume of
nitrogen produced is then measured at room temperature and atmospheric pressure.
(ii) Kjeldahl’s method: a known quantity of nitrogen containing organic compound is
heated with concentrated sulphuric acid. The nitrogen present in the compound is
quantitatively converted into ammonium sulphate. It is then distilled with excess of
sodium hydroxide. The ammonia evolved during this process is passed into a known
volume of H2SO4. The chemical equations involved in the process are
Organic compound Conc H2SO4 → (NH4)2SO4
(NH4)2SO4 + 2NaOH → Na2SO4 + 2NH3 + 2H2O
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2NH3 + H2SO4 → (NH4)2SO4
The acid that is left unused is estimated by volumetric analysis (titrating it against a
standard alkali) and the amount of ammonia produced can be determined. Thus, the
percentage of nitrogen
in the compound can be estimated. This method cannot be applied to the compounds, in
which nitrogen is present in a ring structure, and also not applicable to compounds
containing nitro and azo groups.
23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an
organic compound.
Solution:
Estimation of halogens
Halogens are estimated by the Carius method. In this method, a known quantity of
organic compound is heated with fuming nitric acid in the presence of silver nitrate,
contained in a hard glass tube called the Carius tube, taken in a furnace. Carbon and
hydrogen that are present in the compound are oxidized to form CO2 and H2O
respectively and the halogen present in the compound is converted to the form of AgX.
This AgX is then filtered, washed, dried, and weighed.
Let the mass of organic compound be m g.
Mass of AgX formed= m1 g
1 mol of AgX contains 1 mol of X.
Therefore,
Mass of halogen in m1 g of AgX =Atomic mass of X × m1g
Molecular mass of AgX
Thus, % of halogen will be =Atomic mass of X × m1×100
Molecular mass of AgX × m
Estimation of Sulphur
In this method, a known quantity of organic compound is heated with either fuming nitric
acid or sodium peroxide in a hard glass tube called the Carius tube. Sulphur, present in
the compound, is oxidized to form sulphuric acid. On addition of excess of barium
chloride to it, the precipitation of barium sulphate takes place. This precipitate is then
filtered, washed, dried, and weighed.
Let the mass of organic compound be m g.
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Mass of BaSO4 formed= m1 g
1 mol of BaSO4 = 233 g BaSO4 = 32 g of Sulphur
Therefore, m1 g of BaSO4 contains 32×m1
233g of Sulphur.
Thus, percentage of Sulphur =32×m1×100
233×m.
Estimation of phosphorus
In this method, a known quantity of organic compound is heated with fuming nitric acid.
Phosphorus, present in the compound, is oxidized to form phosphoric acid. By adding
ammonia and ammonium molybdate to the solution, phosphorus can be precipitated as
ammonium phosphomolybdate.
Phosphorus can also be estimated by precipitating it as MgNH4PO4 by adding magnesia
mixture, which on ignition yields Mg2P2O7.
Let the mass of organic compound be m g.
Mass of ammonium phosphomolybdate formed= m1 g
Molar mass of ammonium phosphomolybdate= 1877 g
Thus, percentage of phosphorous =31×m1×100
1877×m%
If P is estimated as Mg2P2O7,
Then, percentage of phosphorous =62×m1×100
222×m%
24. Explain the principle of paper chromatography.
Solution:
In paper chromatography, chromatography paper is used. This paper contains water
trapped in it, which acts as the stationary phase. On the base of this chromatography
paper, the solution of the mixture is spotted. The paper strip is then suspended in a
suitable solvent, which acts as the mobile phase. This solvent rises up the
chromatography paper by capillary action and in the procedure, it flows over the spot.
The components are selectively retained on the paper (according to their differing
partition in these two phases). The spots of different components travel with the mobile
phase to different heights. The paper so obtained (shown in the given figure) is known as
a chromatogram
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25. Why is nitric acid added to sodium extract before adding silver nitrate for testing
halogens?
Solution:
While testing the Lassaigne’s extract for the presence of halogens, it is first boiled with
dilute nitric acid. This is done to decompose NaCN to HCN and Na2S to H2S and to expel
these gases. That is, if any nitrogen and sulphur are present in the form of NaCN and
Na2S, then they are removed. The chemical equations involved in the reaction are
represented as
NaCN+HNO3 → NaNO3 + HCN
Na2S + 2HNO3 → 2NaNO3 + H2S
26. Explain the reason for the fusion of an organic compound with metallic sodium for
testing nitrogen, sulphur and halogens.
Solution:
Nitrogen, sulphur, and halogens are covalently bonded in organic compounds. For their
detection, they have to be first converted to ionic form. This is done by fusing the organic
compound with sodium metal. This is called “Lassaigne’s test”. The chemical equations
involved in the test are
Na+ C + N → NaCN
Na+ S + C + N → NaSCN
2Na+ S → Na2S
Na+ X → NaX
(X = Cl, Br, I)
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Carbon, nitrogen, sulphur, and halogen come from organic compounds.
27. Name a suitable technique of separation of the components from a mixture of calcium
sulphate and camphor.
Solution:
The process of sublimation is used to separate a mixture of camphor and calcium
sulphate. In this process, the sublimable compound changes from solid to vapour state
without passing through the
liquid state. Camphor is a sublimable compound and calcium sulphate is a non-
sublimable solid. Hence, on heating, camphor will sublime while calcium sulphate will
be left behind.
28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its
steam distillation?
Solution:
In steam distillation, the organic liquid starts to boil when the sum of vapour pressure due
to the organic liquid (p1) and the vapour pressure due to water (p2) becomes equal to
atmospheric pressure (p), that is, p = p1 + p2
Since p1 < p2, organic liquid will vapourise at a lower temperature than its boiling point.
29. Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for
your answer.
Solution:
Halogen containing ionic compound+AgNO3 → AgCl (white ppt)
Halogen containing Organic compound AgNO3 → no reaction
In CCl4 the chlorine atoms are covalently bonded with the to carbon so it will not give the
white precipitate of AgCl on heating it with silver nitrate. Cl atom should be present in
ionic formi the compound, it is necessary to prepare the Lassaigne’s extract of CCl4.
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30. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during
the estimation of carbon present in an organic compound?
Solution:
Carbon dioxide is acidic in nature and potassium hydroxide is a strong base.so acid-base
reaction takes place .carbon dioxide reacts with potassium hydroxide to form potassium
carbonate and water
2KOH+ CO2 → K2CO3 + H2O
Thus, the mass of the U-tube containing KOH increases. This increase in the mass of U-
tube gives the mass of CO2 produced. From its mass, the percentage of carbon in the
organic compound can be estimated.
31. Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium
extract for testing sulphur by lead acetate test?
Solution:
CH3COOH(acetic acid ) → CH3COO− + H+
H2SO4(sulphuric acid) → HSO4− +H+
Although the addition of sulphuric acid will precipitate lead sulphate, the addition of
acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate
due to common ion effect. Hence, it is necessary to use acetic acid for acidification of
sodium extract for testing sulphur by lead acetate test.
32. An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being
oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this
substance is subjected to complete combustion.
Solution:
Given: weight of comound = 100 gm
We know : Molecular mass of carbon dioxide, CO2 = 44 g
Percentage of carbon in organic compound= 69 %
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That is, 100 g of organic compound contains 69 g of carbon.
∴ 0.2 g of organic compound will contain =69×0.2
100= 0.138 g of C
Molecular mass of carbon dioxide( CO2) = 12 + 2 × 16 = 44 g
That is, 12 g of carbon is contained in 44 g of CO2.
Therefore, 0.138 g of carbon will be contained in44×0.138
12= 0.506 g of CO2
Thus, 0.506 g of CO2 will be produced on complete combustion of 0.2 g of organic
compound.
Percentage of hydrogen in organic compound is 4.8.
i.e., 100 g of organic compound contains 4.8 g of hydrogen.
Therefore, 0.2 g of the organic compound will contain 4.8×0.2
100= 0.0096 g H
It is known that the molecular mass of water (H2O) is 18 g.
Thus, 2 g of hydrogen is contained in 18 g of water.
∴ 0.0096 g of hydrogen will be contained in of water 18×0.0096
2= 0.0864 g of water.
Thus, 0.0864 g of water will be produced on complete combustion of 0.2 g of the organic
compound.
33. A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method.
The ammonia evolved was absorbed in 60 ml of 0.5 M H2SO4. The residual acid required
60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of
nitrogen in the compound.
Solution:
Given : The total mass of organic compound= 0.50 g
60 mL of 0.5 M solution of NaOH was required by residual acid(H2SO4) for
neutralisation.
=MNaOHVNaOHnNaOH = MH2SO4VH2SO4nH2SO4 (∴ n factor of H2SO4 is 2 )
=60
2 mL of 0.5M H2SO4
= 30 mL of 0.5 M H2SO4
∴ Acid consumed in the absorption of evolved ammonia is (50–30) mL = 20 mL
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Again, 20 mL of 0.5 MH2SO4 = 40 mL of 0.5 MNH3
Also, since 1000 mL of 1 MNH3 contains 14 g of nitrogen,
∴ 40 mL of 0.5 M NH3 will contain14×40
1000× 0.5 = 0.28 g of N
Therefore, percentage of nitrogen in 0.50 g of organic compound=0.28
0.50× 100 = 56 %
Moles of acid= moles of base
⇒ Moles of H2SO4 = moles of NaOH + moles of NH3
⇒ 60 mL × 0.5 M × 2 = 60 mL × 0.5 M + moles of NH3
⇒ 60 mL × 0.5 M × 10−3 × 2 - 60 mL × 0.5 M × 10−3 = moles of NH3
⇒moles of NH3= 60 × 10−3 − 30 × 10−3=30 × 10−3 = 0.03
∴ 1 mole of NH3 contain 14 gram of N
0.03 mole of NH3 contain 14 × 0.03 gram of N = 0.42 gram of N
34. 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius
estimation. Calculate the percentage of chlorine present in the compound.
Solution:
Given that,
Mass of organic compound is 0.3780 g.
Mass of AgCl formed= 0.5740 g
1 mol of AgCl contains 1 mol of Cl.
143.32 gram of AgCl contains 35.5 g chlorine
0.5740 gram of AgCl contains 35.5×0.5740
143.32 gram chlorine = 0.1421 gram chlorine
∴ Percentage of chlorine =weight of Cl
weight of Compound× 100 =
0.1421
0.3780× 100 = 37.59%
Hence, the percentage of chlorine present in the given organic chloro compound is
37.59%.
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35. In the estimation of sulphur by Carius method, 0.468 g of an organic Sulphur compound
afforded 0.668 g of barium sulphate. Find out the percentage of Sulphur in the given
compound.
Solution:
Given : Total mass of organic compound= 0.468 g
Mass of barium sulphate formed= 0.668 g
1 mol of BaSO4=1 mol of sulphur
= 233 g of BaSO4 = 32 g of sulphur
Thus, 0.668 g of BaSO4 contains 32×0.668
233g of sulphur= 0.0917 g of sulphur
Therefore, percentage of sulphur=weight of S
weight of Compound× 100 =
0.0197
0.468× 100 = 19.59 %
Hence, the percentage of sulphur in the given compound is 19.59 %.
36. In the organic compound CH2 = CH–CH2–CH2–C ≡ CH, the pair of hydridised orbitals
involved in the formation of: C2–C3 bond is:
(A) sp– sp2
(B) sp– sp3
(C) sp2– sp3
(D) sp3– sp3
Solution:
In the given organic compound according to IUPAC nomenclature numbering start with
double bond side. the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are
sp2, sp2, sp3, sp3, sp , and sp hybridized respectively. Thus, the pair of hybridized
orbitals involved in the formation of C2 − C3 bond is sp2– sp3.
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37. In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is
obtained due to the formation of:
(A) Na4[Fe(CN)6]
(B) Fe4[Fe(CN)6]3
(C) Fe2[Fe(CN)6]
(D) Fe3[Fe(CN)6]4
Solution:
In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is
boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process,
sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate
(II). Then, on heating with
sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which
is Prussian blue in colour. The chemical equations involved in the reaction can be
represented as
6CN− + Fe2+ → [Fe(CN)6]4−
Hence, the Prussian blue colour is due to the formation of Fe4[Fe(CN)6]3.
38. Which of the following carbocation is most stable?
(A) (CH3)3C. C+
H2
(B) (CH3)3C+
(C) CH3CH2C+
H2
(D) CH3C+
H CH2CH3
Solution:
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(CH3)3C+
is a tertiary carbocation. A tertiary carbocation is the most stable carbocation
due to the electron releasing effect of three methyl groups. An increased +I effect by
three methyl groups stabilizes the positive charge on the carbocation.
39. The best and latest technique for isolation, purification and separation of organic
compounds is:
(A) Crystallisation
(B) Distillation
(C) Sublimation
(D) Chromatography
Solution:
Chromatography is the most useful and the latest technique of separation and purification
of organic compounds. It was first used to separate a mixture of coloured substances.
40. The reaction:
CH3CH2I + KOH(aq) → CH3CH2OH+ KI
is classified as:
(A) electrophilic substitution
(B) nucleophilic substitution
(C) elimination
(D) addition
Solution:
CH3CH2(l) + KOH ((aq)) → CH3CH2OH+ Kl
It is an example of nucleophilic substitution reaction. The hydroxyl group of KOH (OH–) with a lone pair of itself acts as a nucleophile and substitutes iodide ion in CH3CH2I to
form ethanol.
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